[Swift]LeetCode1224. 最大相等频率 | Maximum Equal Frequency

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Given an array nums of positive integers, return the longest possible length of an array prefix of nums, such that it is possible to remove exactly one element from this prefix so that every number that has appeared in it will have the same number of occurrences.

If after removing one element there are no remaining elements, it's still considered that every appeared number has the same number of ocurrences (0).

Example 1:

Input: nums = [2,2,1,1,5,3,3,5]
Output: 7
Explanation: For the subarray [2,2,1,1,5,3,3] of length 7, if we remove nums[4]=5, we will get [2,2,1,1,3,3], so that each number will appear exactly twice.
Example 2:

Input: nums = [1,1,1,2,2,2,3,3,3,4,4,4,5]
Output: 13
Example 3:

Input: nums = [1,1,1,2,2,2]
Output: 5
Example 4:

Input: nums = [10,2,8,9,3,8,1,5,2,3,7,6]
Output: 8

Constraints:

2 <= nums.length <= 10^5
1 <= nums[i] <= 10^5

---

给出一个正整数数组 nums，请你帮忙从该数组中找出能满足下面要求的 最长 前缀，并返回其长度：

从前缀中 删除一个 元素后，使得所剩下的每个数字的出现次数相同。
如果删除这个元素后没有剩余元素存在，仍可认为每个数字都具有相同的出现次数（也就是 0 次）。 

示例 1：

输入：nums = [2,2,1,1,5,3,3,5]
输出：7
解释：对于长度为 7 的子数组 [2,2,1,1,5,3,3]，如果我们从中删去 nums[4]=5，就可以得到 [2,2,1,1,3,3]，里面每个数字都出现了两次。
示例 2：

输入：nums = [1,1,1,2,2,2,3,3,3,4,4,4,5]
输出：13
示例 3：

输入：nums = [1,1,1,2,2,2]
输出：5
示例 4：

输入：nums = [10,2,8,9,3,8,1,5,2,3,7,6]
输出：8

提示：

2 <= nums.length <= 10^5
1 <= nums[i] <= 10^5

---

Runtime: 440 ms

Memory Usage: 23.6 MB

class Solution {
    func maxEqualFreq(_ nums: [Int]) -> Int {
        var count:[Int] = [Int](repeating:0,count:100001)
        var freq:[Int] = [Int](repeating:0,count:100001)
        var res:Int = 0
        let N:Int = nums.count
        var a:Int = 0
        var c:Int = 0
        var d:Int = 0
        for n in 1...N
        {
            var a:Int = nums[n - 1]
            freq[count[a]] -= 1
            count[a] += 1
            c = count[a]
            freq[count[a]] += 1
            
            if freq[c] * c == n && n < N
            {
                res = n + 1
            }
            d = n - freq[c] * c
            if (d == c + 1 || d == 1) && freq[d] == 1
            {
                res = n
            }
        }
        return res
    }
}

---

444ms

class Solution {
    func maxEqualFreq(_ nums: [Int]) -> Int {
        var count = [Int](repeating: 0, count: 100001)
        var freq = [Int](repeating: 0, count: 100001)
        
        let N = nums.count
        
        var res = 0
        var a = 0
        var c = 0
        var d = 0
        
        for n in 1...N {
            a = nums[n-1]
            freq[count[a]] -= 1
            count[a] += 1
            c = count[a]
            freq[count[a]] += 1
            
            if freq[c] * c == n && n < N {
                res = n + 1
            }
            d = n - freq[c] * c
            if (d == c + 1 || d == 1) && freq[d] == 1 {
                res = n
            }
        }
        
        return res
    }
}

---

492ms

class Solution {
    func maxEqualFreq(_ nums: [Int]) -> Int {
        let MAX_NUM = 1000_005
        var freq = [Int](repeating: 0, count: MAX_NUM)
        var freq_freq = freq
        var ans = 0
        let N = nums.count
        for i in nums.indices {
            let tmp = nums[i]
            freq_freq[freq[tmp]] -= 1

            freq[tmp] += 1
            let c = freq[tmp]
            freq_freq[c] += 1

            if freq_freq[c]*c == i+1 && i+1<N {
                ans = i + 1 + 1
            }

            let d = i + 1 - freq_freq[c] * c
            if (d == c + 1 || d == 1) && freq_freq[d] == 1 {
                ans = i + 1
            }
        }
        return ans
    }
}

---

528ms

class Solution {
    func maxEqualFreq(_ nums: [Int]) -> Int {
        let numCount = nums.count
        var countHash = [Int:Int]()
        var freqHash = [Int:Int]()
        var result = 0
        
        for i in 0..<numCount {
            let num = nums[i]
            
            countHash[num] = (countHash[num] ?? 0) + 1
            let freq = countHash[num]!
            freqHash[freq] = (freqHash[freq] ?? 0) + 1
            let freqCount = freqHash[freq]! * freq 
            
            if freqCount == i {
                // All same frequency at i (which is 0 based), there's one we can remove
                result = max(result, i+1)  
            } else if freqCount == i + 1, i != numCount - 1 { 
                // print(i)
                // print(freqHash)
                // All same frequency, and still not at the last character. Add one more 
                result = max(result, i+2)  
            }
        }
        
        return result
    }
}

---

812ms

class Solution {
    
    var dict = [Int: Int]()  // 频率 --> 个数
    var m = [Int: Int]()  // 当前数的频率
    
    func maxEqualFreq(_ nums: [Int]) -> Int {
        var result = 1 
        for i in 0..<nums.count {
            
            m[nums[i], default: 0] += 1 
            let last = m[nums[i]]! - 1
            
            //维护前缀
            if let counter = dict[last] {
                dict[last] = counter - 1
                if counter == 1 {
                    dict.removeValue(forKey: last)
                }
            }
            if let freq = m[nums[i]] {
                dict[freq, default:0] += 1
            }
            if check() { result = i + 1 }
        }
        return result
    }
    

    fileprivate func check() -> Bool {
        
        //只出现一个数字
        if m.count == 1 { return true }  

        // 出现很多数字，但是都次数都是1
        if dict.count == 1 { return dict[1] != nil }

        if dict.count != 2 { return false }
        
        //只有一个数字出现一次，其他都出现多次，且次数相等
        for (freq, counter) in dict {
            if freq == 1 && counter == 1 { return true }
        }
        
        //所有数字都出现多次，但是大多数次数都相等，只有一个特殊多了一次
        let keys = Array(dict.keys)
        if let counter0 = dict[keys[0]], let counter1 = dict[keys[1]] {
            if keys[0] - keys[1] == 1 { return counter0 == 1 }
            if keys[1] - keys[0] == 1 { return counter1 == 1 }
        }
        return false
    }
}